Question: What is the value of the following logarithm? $\log_{8} \left(\dfrac{1}{8}\right)$
If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $8^{y} = \dfrac{1}{8}$ Any number raised to the power $-1$ is its reciprocal, so $8^{-1} = \dfrac{1}{8}$ and thus $\log_{8} \left(\dfrac{1}{8}\right) = -1$.